Should be easy for Mathematicians. The partial deferential equation is from Thermodynamics and the book presents the following without a worked sequence. I have taken Calculus 3 and differential equations (introduction to ordinary differential equations).

$G=H+(\partial G/\partial T)T$

"can be rewritten as"

$\partial /\partial T(G/T)=-H/T^2$

Can someone show me the work? I have tried to separate variables and integrate but I get

$\dfrac{1}{G-H}\partial G=\dfrac{\partial T}{T}$

Integrating the left gives a logarithm so I know this can't be the correct procedure.

Thanks, Mathematicians.

  • $\begingroup$ Remeber:the derivative of $G/T$ is the derivative of the product $G\cdot 1/T$. $\endgroup$ – anderstood Jan 30 '15 at 4:43

By the quotient rule,

$$\frac{\partial}{\partial T}(G/T) = \frac{T \frac{\partial G}{\partial T} - G \frac{\partial T}{\partial T}}{T^2} = \frac{\partial G \over \partial T}{T} - \frac{G}{T^2}$$

Hence $\displaystyle \frac{\partial}{\partial T}(G/T)=-H/T^2$ is equivalent to

$$T \frac{\partial G}{\partial T} - G = -H$$

which is equivalent to your original expression.


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